Show $f_{n}(x) = n \sin(\frac{x}{n})$ converges point wise on the compact set $[-R,R]$

Question

Show $f_{n}(x) = n \sin(\frac{x}{n})$ converges point wise on the compact set $[-R,R]$

Playing with the function I was able to deduce that $f_{n}(x) \rightarrow f(x) = x$.

My question is how to show this formally?

I know that to show point wise convergence, I have to show that for every $\epsilon > 0$ and for all $x \in [-R,R]$ there exists a $N > 0 $ s.t $|n \sin(\frac{x}{n}) - x| < \epsilon$.

In situations that don't have a trigonometric function you could possibly solve for the $N$ explicitly, but that won't work in this situation. So my question is what to do in a situation like this? For this question I know that locally $\sin(x) \sim x$. That being the case would this count towards a solution?

$$|n\sin(\frac{x}{n}) - x| \leq |n(\frac{x}{n}) - x| = 0 < \epsilon$$

Thoughts and feedback would be appreciated.

Answer 1

Write $$n \sin \left(\frac{x}{n}\right)=x \left(\frac{\sin \left(\frac{x}{n}\right)}{\frac{x}{n}}\right)$$ and use

$$\left(\frac{\sin \left(\frac{x}{n}\right)}{\frac{x}{n}}\right) \to 1$$ as $n \to \infty$

That is, $$ |n \sin \left(\frac{x}{n}\right)-x |=x \left|\left(\frac{\sin \left(\frac{x}{n}\right)}{\frac{x}{n}}\right) -1 \right| <x \left(1-\cos \frac{x}{n}\right)=x \cdot 2 \sin^2 \frac{\frac{x}{n}}{2} < x \cdot \frac{(\frac{x}{n})^2}{2}=\frac{x^3}{2n^2}$$ Choose suitable $n$ from this inequality!

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Here I use the following facts :

• Since $\cos x<\frac{\sin x}{x}<1$ $$\bigg|\frac{\sin x}{x}-1\bigg|<1-\cos x$$
• $1-\cos x=2\sin^2\frac{x}{2}\leqslant\frac{x^2}{2}$